MAE237D_HW1

MAE237D_HW1 - Giacomo Po ID 803055024 MAE 237D Homework 1...

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Unformatted text preview: Giacomo Po ID 803055024 MAE 237D Homework 1 Problem 1 From tables, the atomic mass of H 3 is 3 . 016049 amu , therefore one mole is 3 . 016049 g . This means that 1 kg of tritium contains: n t = 1000 3 . 016049 . 6022137 10 24 = 1 . 997 10 26 Problem 2 Considering the thermal efficiency the thermal power is P = 400 MW . 4 = 1000 MW The total energy consumed in a year is: E = 1000 365 24 3600 = 3 . 1536 10 16 J The thermal energy produced by a fusion tritium is 17 . 58 MeV and therefore the number of tritium moles necessary to produce the total thermal energy is: n = 3 . 1536 10 16 J 17 . 58 1 . 602 10- 13 J moles . 6022137 10 24 = 0 . 18594 10 5 moles Since one mole is 3 . 016049 g then the previous number of moles corresponds to a mass of: m = 3 . 016049 . 18594 10 5 = 0 . 560804151 10 5 g 56 . 08 kg Problem 3 For the reaction 3 1 T + 2 1 D 4 2 He + n : Q = [(3 . 016049 + 2 . 014102)- (4 . 002604 + 1 . 008665)] 931 MeV amu = 17 . 58 MeV For the reaction 2 1 D + 2 1 D 3 2 He + n : Q = [(2 . 014102 + 2 . 014102)- (3 . 016 + 1 . 008665)] 931 MeV amu = 3 . 29 MeV For the reaction...
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This note was uploaded on 05/11/2008 for the course MAE 237d taught by Professor Bb during the Spring '08 term at UCLA.

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MAE237D_HW1 - Giacomo Po ID 803055024 MAE 237D Homework 1...

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